EXAM 3 REVIEW LBS 172 REACTION MECHANISMS

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1 EXAM 3 REVIEW LBS 172 REACTION MECHANISMS GENERAL -Step by step process of bond making and breaking by which reactants become products -Summation of steps must be equal to overall reaction -Example: NO 2 (g) + F 2 (g) FNO 2 + F(g) NO 2 (g) + F(g) FNO 2 (g) Overall: 2NO 2 (g) + F 2 (g) 2FNO 2 (g) TERMS -Elementary reaction: individual step in a mechanism; one chemical event Ex (from above rxn): NO 2 (g) + F 2 (g) FNO 2 + F(g) -Intermediate: species that is formed and subsequently used up in the reaction Ex: F(g) -Molecularity: number of particles colliding in an elementary step 1- unimolecular 2- bimolecular 3- termolecular Ex: NO(g) + F(g) FNO 2 (g) is a bimolecular step -Catalyst: species that is used up by the mechanism and is subsequently reformed RATES FOR ELEMENTARY STEPS -Rate equation of an elementary step is based on reaction stoichiometry -Rate = product of rate constant and the concentrations of each reacting particle (Neither solids nor pure liquid reactants are used in rate equations) For any elementary step: A + B C + D, rate = k[a][b] Ex: Rate for step NO 2 (g) + F 2 (g) FNO 2 + F(g) = k[no 2 ][F 2 ] RATE EQUATIONS BASED ON REACTION MECHANISMS -The rate of any reaction is limited by its slowest step, the rate of which is often nearly the same as the overall reaction rate -Rate determining step (RDS): the slowest elementary step in a reaction mechanism -The RDS involves the highest total energy of all elementary steps in a given mechanism -The rate of a rate determining step is assumed to be equal to the rate of its overall reaction 1. A + B X + M, Slow, large E a 2. M + A Y, Fast, small E a Overall: 2A + B X + Y Step 1 is RDS, so rate of overall reaction = k step 1 [A][B]

2 -In a laboratory setting, proposed mechanisms can be disproved by examining rate dependence on reactants in RDS. Mechanisms cannot, however, be proved correct. -Reaction mechanisms involving an equilibrium step Common case: a fast equilibrium step followed by a slow step Ex: 1. NO 2 (g) NO(g) + O(g) fast, equilibrium, where k 1 is constant for forward rxn and k -1 is constant for reverse reaction 2. O(g) + NO 2 (g) NO(g) + O 2 (g) slow, k 2 is constant Overall: 2NO 2 (g) 2NO(g) + O 2 (g) -Rate equation may be written as: k 2 [O][NO 2 ] -O is an immeasurable intermediate, and should not appear in the rate equation -Since in equilibrium, k 1 [NO 2 ] = k -1 [NO][O], [O] can be written in terms of measurable species [O] = k 1 [NO 2 ] / (k -1 [NO]) -Rate equation can be written as: k 2 k 1 [NO 2 ] / (k -1 [NO]) [NO 2 ] COLLISION THEORY AND ACTIVATION ENERGY -Collision theory states that for any reaction to occur, three conditions must be met: 1. Reacting molecules must collide 2. Collision must have enough energy to break bonds 3. Molecules must collide in the appropriate orientation -Activation energy (E A ): the minimum amount of energy that must be absorbed by a system to cause it to react -The proportion of molecules in a sample that have energy at or above E A is directly related to the speed of their reaction -Reaction coordinate diagram Shows chemical potential energy over time during a reaction Ex: P O T E N T I A L E A Reactants E A E N E R G Y TIME Products

3 The energy of the reactants at left is higher than the energy of the products at right, therefore the reaction is exothermic (reactants would have lower E than products if endothermic) E A is the difference from the beginning E of the sample and the energy required for the sample to react When the sample has reached E A, there is enough energy to break the appropriate bond(s), this occurs at the peak. A single peak represents one step New bonds are created to form products. To begin a reverse reaction, E A must be added to the products to reach the peak -Effects of Temperature Heating a sample has the effect of increasing the fraction of molecules in a sample having higher energies. This in turn increases the fraction of molecules at or above the activation energy barrier and therefore increases the rate of reaction Heat does not change concentration of reactants, so for the rate to increase, an increase in temperature must increase the rate constant Arrhenius equation: rate constant k = Ae -Ea/(RT), where A is a parameter called the frequency factor, R is the gas constant, E A is activation energy, and T is temperature EQUILIBRIUM EQUILIBRIUM CONSTANT AND REACTION QUOTIENT -Equilibrium constant Reaction: aa + bb cc + dd at equilibrium Equilibrium constant K eq = ([C] c [D] d )/([A] a [B] b ) K eq may only be expressed at equilibrium K eq depends only on the reaction and temperature K eq has no units Solid and liquid reactants and products are omitted in K eq calculations K eq may be written as K c, as it is based on concentrations From equation for K eq, if K eq << 1, then reaction is reactant favored; if K eq >> 1, then reaction is product favored -Equilibrium constant can be written in terms of partial pressures for reactions involving gasses Reaction: aa(g) + bb(g) cc(g) + dd(g) at equilibrium K p = (P C c * P D d ) / (P A a * P B b ) By the ideal gas law, K p = K c (RT) n, where n = change in moles of gas -Reaction quotient, Q Reaction: aa + bb cc + dd Q = ([C] c [D] d )/([A] a [B] b ) Q is the same expression as K eq, but it may be expressed when a system is not in equilibrium. System is at equilibrium if Q = K eq. Q can be used to predict concentration changes as systems approach equilibrium

4 -If Q < K, reaction will shift right, converting more reactants into products to reach equilibrium -If Q > K, reaction will shift left, converting more products into reactants Q can be applied similarly to partial pressures, and compared to K p -LeChatelier s Principle and rates Equilibrium will react to a stress to reach equilibrium This can be explained with rates Ex: Co Cl - CoCl 4 2-, If Cl - is added, Q will decrease, rxn will make more CoCl 4 2- to reach equilibrium. Why? At equilibrium: k[cl - ] 4 [Co 2+ ] = k[cocl 4 2- ], if Cl - is added, the rate of the forward reaction increases, increasing [CoCl 4 2- ] and then increasing the reverse reaction to stabilize at equilibrium In endothermic reactions, the addition of heat has the same effect on rate as the addition of reactant In exothermic reactions, the addition of heat has the same effect on rate as the addition of product EQUILIBRIUM CALCULATIONS -General steps to solve equilibrium problems 1. Write out equation 2. Use Q to determine shift 3. Set up ICE table, use balanced equation to find ratios 4. Plug in Eq values into K expression 5. Solve for variable change 6. Use change to find equilibrium concentrations 7. Check answer -To avoid using the quadratic formula, the addition or subtraction of a variable change (x) in a K expression can be neglected only if the addition/subtraction of x does not significantly affect the number that is being added to/subtracted from. -How to use an ICE table Ex: C(graphite) + CO 2 (g) 2CO(g). The system was kept at constant volume at 1000K until equilibrium was reached. At this temperature, K = The initial concentration of CO 2 was 0.012M. Calculate the equilibrium concentration of CO. C(graphite) + CO 2 (g) 2CO(g) Initial Change - -x + 2x Equilibrium x 2x With these assumptions, we can set K = = (2x) 2 / (0.012-x). If the x in the denominator is neglected, then x is calculated as 0.008, which significantly effects the concentration of CO 2. So the quadratic formula must be used. x is correctly calculated to be The equilibrium concentration of CO then is calculated to be 2(0.0057) = M.

5 ACIDS AND BASES ACIDS BASES AND THEIR REACTIONS -Bronsted definitions: Acid: a substance that can donate a proton to any other substance Base: a substance that can accept a proton from any other substance -Reactions Involve the transfer of a proton (H + ) Acid + Base Conjugate Base + Conjugate Acid Acids and their conjugate bases differ only by the presence of one proton, as do bases and their conjugate acids Ex: HCO 3 - (aq) + H 2 O(l) H 3 O + (aq) + CO 3 2- (aq) Acid Base Conj. Acid Conj. Base The acid HCO 3 - and its conjugate base CO 3 2- differ by only one H + The base H 2 O and its conjugate acid H 3 O + differ by only one H + Ex: NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH - (aq) Base Acid Conj. Acid Conj. Base -Strong Acids HCl, HBr, HNO 3, HI, H 2 SO 4, HClO 4 These create a complete forward reaction, not an equilibrium reaction. The conjugate bases are relatively too weak to induce reverse reactions. Ex: HCl(aq) + H 2 O H 3 O + (aq) + Cl - (aq), Cl - is too weak of a base to pull a proton from H 3 O + in this system, so the reaction is one way All other (weak) acids create bases with sufficient strength to maintain equilibrium. The stronger the acid, the weaker its conjugate base. -Strong Bases NaOH, KOH, LiOH, Ba(OH) 2, H -, RO -, R -, NH 2 - These create a complete forward reaction, not an equilibrium reaction. The conjugate acids are relatively too weak to induce reverse reactions. Ex: H - + H 2 O H 2 + OH - (aq), H 2 has nearly no acid strength, so the reaction is one way All other (weak) bases create acids with sufficient strength to maintain equilibrium. The stronger the base, the weaker its conjugate acid. -Monoprotic, Polyprotic, and Amphiprotic Monoprotic acids/bases: Bronstead acids/bases that are capable of donating/accepting one proton Ex: acid: HCl, base: NaOH Polyprotic acids/bases: Bronstead acids/bases that are capable of donating/accepting more than one proton Ex: acid: H 2 SO 4, base: CO 3 2- Amphiprotic: describes species that can behave either as Bronstead acids or bases Ex: Water as base: H 2 O(l) + HCl(aq) H 3 O + (aq) + Cl - (aq) Water as acid: H 2 O(l) + NH 3 (aq) NH 4 + (aq) + OH - (aq)

6 -Lewis Acids Electron acceptors that, although do not donate H + themselves, increase the acidity of solution. Transition metal cations are Lewis acids. Ex: Fe 2+, as in hard water. Fe 2+ attracts 6 H 2 O molecules in a octahedral pattern. O-H bonds in water weaken due to the pulling of e - by Fe 2+. H + is then more easily removed by a base, resulting in a more acidic solution. ACID AND BASE MEASURES - 2H 2 O H 3 O + (aq) + OH - (aq) K w = [H 3 O + ][OH - ]. At 25 C, K w = 1*10-14 for all aqueous solutions - ph = -log[h 3 O + ] = -log[h + ] - poh = -log[oh - ] - ph + poh = 14 for all aqueous solutions - Solution is acidic if: [H 3 O + ] > [OH - ], ph < 7, poh > 7 - Solution is basic if: [H 3 O + ] < [OH - ], ph > 7, poh < 7 - Solution is neutral if: [H 3 O + ] = [OH - ] = 1*10-7 M, ph = poh = 7 EQUILIBRIUM CONSTANTS FOR ACIDS AND BASES -For a general acid HA: HA(aq) + H 2 O(l) H 3 O + + A - (aq) K a = ([H 3 O + ][A - ]) / [HA] K a is directly related to acid strength -For a general weak base B: B(aq) + H 2 O(l) BH + (aq) + OH - (aq) K b = ([BH + ][OH - ]) / [B] K b is directly related to base strength - K a * K b = K w for all aqueous solutions -Logarithmic measures pk a = -logk a pk a is indirectly related to acid strength pk b = -logk b pk b is indirectly related to base strength PREDICTING ph OF A SALT SOLUTION -Consider the ions of the salt -Conjugate bases of weak acids are basic, while those of strong acids are neutral -Conjugate acids of weak bases are acidic, while those of strong based are neutral -Alkali metal cations have no effect on ph -Transition metal cations are acidic (Lewis acids) -Ex: NaNO 3 ions: Na + has no effect on ph. NO 3 - is the conjugate base of HNO 3, a strong acid, therefore it is neutral. So NaNO 3 solution is neutral. EQUILIBRIUM CALCULATIONS -Use the same techniques as non-acid/base equilibrium calculations discussed above -Use given ph and/or poh information to find [H 3 O + ] and/or [OH - ]

7 -Use K w = K a * K b to convert between K a and K b -In problems with polyprotic acids or bases, a full cycle of calculations are necessary for each donation/acceptance of a proton. E.g., three ICE tables will be necessary to find end concentrations of all species in a reaction with a triprotic acid. BUFFERS -Buffer: a solution that resists ph changes -Composed of a weak acid and a weak base for conjugate acid/base pair -Most effective when ph pk a, generic: [A - ] [HA] -Buffer capacity: measure of how much acid or base a buffer can absorb without significant ph changes; affected by total amount of A - and HA present -Finding ph of a buffer (2 options) 1. Do an equilibrium calculation 2. Use Henderson Hasselbach equation ph = pk a + log([a - ]/[HA]) poh = pk b + log([hb + ]/[B]) Only can be used when the variable change (x) in the equilibrium calculation is insignificant